Answer to Question #233816 in Statistics and Probability for Vrushank

Question

From the following uniformly distributed data scene calculate the probability w.r.t statement.

The weekly output of a steel mill is a uniformly distributed random variable that lies between 110 and 175 metric tons.

1. Compute the probability that the steel mill will produce more than 150 metric tons next week.

2. Determine the probability that the steel mill will produce between 120 and 160 metric tons

Answer 1

The probability dense function is $f_X(x)=1/(175-110)=1/65$ for $x\in[110,175]$ and $f_X(x)=0$ for another $x$.

$P(X\geq 150)=\int\limits_{150}^{175}f_X(x)dx=\int\limits_{150}^{175}\frac{1}{65}dx=5/13$

$P(120\leq X\leq 160)=\int\limits_{120}^{160}f_X(x)dx=\int\limits_{120}^{160}\frac{1}{65}dx=8/13$

Answer. 1) 5/13; 2) 8/13